15. Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1

Ask & Answer Question

Application of Integrals Ncert Solutions Maths class 12th Maths Class 12th Application of Integrals

15. Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1

3 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

The equation of the given circle is

$= x^{2} + y^{2} = 1$ - (1)

$= (x-1) 2 + y^{2} = 1$ - (1) - (2)

Equation (1) is a circle with centre 0 (0,0) and radius 1. Equation (2) is a circle with centre c (1,0) and radius 1.

Solving (1) and (2)

0 Upvotes 0 Downvotes

Similar Questions for you

Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then a is equal to………….

Raj Pandey

$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ differentiating w.r.to x

$x^{3} f (x) + 3 x^{2} \frac{f^{3} (x)}{x^{3}} = 3 f^{2} (x) f' (x) \Rightarrow 3 y^{2} \frac{d y}{d x} = x^{3} y = \frac{3 y^{3}}{x} \Rightarrow 3 x y \frac{d y}{d x} = x^{4} + 3 y^{2}$

After solving we get $y^{2} = \frac{x^{4}}{3} + c x^{2}$ also curve passes through (3, 3) Þ c = -2

$∴ y^{2} = \frac{x^{4}}{3} - 2 x^{2}$ which passes through $(α, 6 \sqrt[]{10})$ $∴ \frac{α^{4} - 6 α^{2}}{3} = 360 \Rightarrow α = 6$

The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y^a is $\frac{364}{3},$ is equal to:

Raj Pandey

Since a is a odd natural number then $| \int_{1}^{3} y^{a} d y | = \frac{364}{3} \Rightarrow | {(\frac{y^{a + 1}}{a + 1})}_{1}^{3} | = \frac{364}{3} \Rightarrow \frac{3^{a + 1}}{a + 1} = \frac{364}{3}$